적분의 활용 전체 문제

By definition, the average is

$$\begin{aligned} \frac{1}{e-1}\int_1^e x^2\log x\ \mathrm{d}x & \\ \text{To antidifferentiate, we use integration by parts with $u=\log x$ and $\mathrm{d}v=x^2\,\mathrm{d}x$, hence $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$ and $v=\frac{1}{3}x^3$.} \\ \frac{1}{e-1}\int_1^e x^2\log x\ \mathrm{d}x&= \frac{1}{e-1}\left(\left[\frac{1}{3}x^3\log x\right]_1^e - \int_1^e \frac{1}{3}x^2\,\mathrm{d}x \right) \\&= \frac{1}{e-1}\bigg[\frac{x^3}{3}\log x-\frac{x^3}{9}\bigg]_{x=1}^{x=e}\\ &=\frac{1}{e-1}\bigg[\frac{e^3}{3}-\frac{e^3}{9}+\frac{1}{9}\bigg]\\ &=\frac{1}{e-1}\bigg[\frac{2}{9}e^3+\frac{1}{9}\bigg] \end{aligned}$$

By definition, the average value in question equals

$$\frac1{\pi/2-0} \int_0^{\pi/2} ( 3\cos^3x + 2\cos^2x ) \,\mathrm{d}x = \frac2\pi \bigg( \int_0^{\pi/2} 3\cos^3x \,\mathrm{d}x + \int_0^{\pi/2} 2\cos^2x \,\mathrm{d}x \bigg)$$

For the first integral we use the substitution $u = \sin x$, $\mathrm{d}u=\cos x\,\mathrm{d}x$, $\cos^2 x=1-\sin^2 x = 1-u^2$. Note that the endpoints $x=0$ and $x=\frac\pi2$ become $u=0$ and $u=1$, respectively.

$$\begin{aligned} \int_0^{\pi/2} 3 \cos^3x \,\mathrm{d}x &= \int_0^{\pi/2} 3 \cos^2 x \cos x \,\mathrm{d}x \\ &= \int_0^{1} 3(1-u^2) \,\mathrm{d}u \\ &= (3u-u^3)\Big|_0^1 = 2. \end{aligned}$$

For the second integral we use the trigonometric identity $\cos^2x\,\mathrm{d}x = \frac{1 + \cos(2x)}{2}$.

$$\begin{aligned} 2 \int_0^{\pi/2} \cos^2x\,\mathrm{d}x &= \int_0^{\pi/2} \big(1 + \cos(2x) \big) \,\mathrm{d}x \\ &= \bigg[ x + \frac{1}{2}\sin(2x) \bigg]_0^{\pi/2} = \frac\pi2 \end{aligned}$$

Therefore, the average value in question is

$$\frac2\pi \bigg( \int_0^{\pi/2} 3\cos^3x \,\mathrm{d}x + \int_0^{\pi/2} 2\cos^2x \,\mathrm{d}x \bigg) = \frac2\pi \bigg( 2 + \frac\pi2 \bigg) = \frac{4}{\pi}+1.$$

By definition, the average value in question equals

$$\text{Ave}=\frac1{\pi/k-0} \int_0^{\pi/k} \sin(kx) \,\mathrm{d}x$$

To evaluate the integral, we use the substitution $u = kx$, $\mathrm{d}u=k\,\mathrm{d}x$. Note that the endpoints $x=0$ and $x=\pi/k$ become $u=0$ and $u=\pi$, respectively. So

$$\begin{aligned} \text{Ave}=\frac{k}{\pi} \int_0^\pi \sin(u) \,\frac{\mathrm{d}u}{k} =\frac{1}{\pi}\Big[-\cos(u)\Big]_0^\pi =\frac{2}{\pi} \end{aligned}$$

Remark: the average does not depend on $k$. To see why this is, note that $\sin(kx)$ runs between $-1$ and $1$ as $x$ changes. When $x=0$, $kx=0$, and when $x=\pi/k$, $kx=\pi$. So, our function $\sin (kx)$ runs exactly from $\sin 0 =0$ to $\sin(\pi/2)=1$, then back down to $\sin\pi=0$.

By definition, the average temperature is $$\frac{1}{3}\int_0^3 T(x)\,\mathrm{d}x =\frac{1}{3}\int_0^3 \frac{80}{16-x^2}\,\mathrm{d}x$$ We don't see an obvious substitution, but integrand is a rational function. The degree of the numerator is strictly less than the degree of the denominator, so we factor the denominator and use a partial fraction decomposition.

$$\begin{aligned} \frac{80}{16-x^2}&=\frac{80}{(4-x)(4+x)} = \frac{A}{4-x}+\frac{B}{4+x}\\ 80&=A(4+x)+B(4-x) \\ \text{ Setting $x=4$, we see $80=8A$, so $A=10$. Setting $x=-4$, we see $80=8B$, so $B=10$. } \\ \frac{1}{3}\int_0^3 \frac{80}{16-x^2}\,\mathrm{d}x&=\frac{1}{3}\int_0^3 \frac{80}{(4-x)(4+x)}\,\mathrm{d}x=\frac{1}{3}\int_0^3 \Big[\frac{ 10}{4-x}+\frac{ 10}{4+x}\Big]\,\mathrm{d}x\\ &=\frac{1}{3}\int_0^3 \Big[-\frac{10}{x-4}+\frac{10}{4+x}\Big]\,\mathrm{d}x\\ &=\frac{10}{3}\Big[-\log|x-4|+\log|x+4|\Big]_0^3 \\ &=\frac{10}{3}\log\Big|\frac{x+4}{x-4}\Big|\bigg|_0^3 =\frac{10}{3}[\log 7-\log 1] \\ &=\frac{10}{3}\log 7\quad\text{ degrees Celsius} \end{aligned}$$

By definition, the average value is

$$\begin{aligned} \frac{1}{e-1}\int_1^e\frac{\log x}{x}\ \mathrm{d}x& \\ \text{To integrate, we use the substitution $u=\log x$, $\mathrm{d}u=\frac{1}{x}\,\mathrm{d}x$. Then the limits of integration become 0 and 1, respectively.} \\ \frac{1}{e-1}\int_1^e\frac{\log x}{x}\ \mathrm{d}x& =\frac{1}{e-1}\int_0^1u\ \mathrm{d}u =\frac{1}{e-1}\bigg[\frac{u^2}{2}\bigg]_0^1 =\frac{1}{2(e-1)} \end{aligned}$$

By definition, the average value is:

$$\begin{aligned} \frac{1}{2\pi}\int_0^{2\pi}\cos^2x\,\mathrm{d}x =\frac{1}{2\pi}\cdot\frac{1}{2}\int_0^{2\pi}\big(\cos(2x)+1\big)\,\mathrm{d}x =\frac{1}{4\pi}\Big[\frac{\sin(2x)}{2}+x\Big]_0^{2\pi} =\frac{1}{4\pi}\cdot 2\pi =\frac{1}{2} \end{aligned}$$

Before we start answering questions, let's look at our function a little more carefully. The term $50\cos\left(\frac{t}{12}\pi\right)$ has a period of 24 hours, while the term $200\cos\left(\frac{t}{4380}\pi\right)$ has a period of one year. So, the former term describes a standard daily variation, while the latter gives a seasonal variation over the year.

(a) Using the definition of an average, the average concentration over one year ($t=0$ to $8760$) is:

$$\begin{aligned} &\frac{1}{8760}\int_0^{8760}\left(400+ 50\cos\left(\frac{t}{12}\pi\right)+ 200\cos\left(\frac{t}{4380}\pi\right)\right)\,\mathrm{d}t\\ &=\frac{1}{8760}\int_0^{8760}400\,\mathrm{d}t + \frac{50}{8760}\int_0^{8760}\cos\left(\frac{t}{12}\pi\right)\,\mathrm{d}t + \frac{200}{8760}\int_0^{8760}\cos\left(\frac{t}{4380}\pi\right)\,\mathrm{d}t\\ &= 400 + \frac{5}{876}\left[\frac{12}{\pi}\sin\left(\frac{t}{12}\pi\right)\right]_0^{8760}+ \frac{5}{219}\left[\frac{4380}{\pi}\sin\left(\frac{t}{4380}\pi\right)\right]_0^{8760} \\ \text{Since $\frac{8760}{12} = 730$, which is even, $\sin\left(\frac{8760}{12}\pi\right)=\sin(0)=0$. Also, $\sin\left(\frac{8760}{4380}\pi\right)=\sin(2\pi)=0$. } \\ &=400+ \frac{5}{876}(0) + \frac{5}{219}(0)\\ &=400\quad\text{ ppm} \end{aligned}$$

Remark: for the portions of the integral in red and blue, we also could have noticed that the integrand goes through a whole (integer) number of periods. For every period, the net signed area between the curve and the $x$-axis is zero, so we could have seen from the very beginning these terms would contribute 0 to the final average.

(b) Using the definition of an average, the average concentration over the first day ($t=0$ to $t=24$) is:

$$\begin{aligned} &\frac{1}{24}\int_0^{24} \left(400+ 50\cos\left(\frac{t}{12}\pi\right)+ 200\cos\left(\frac{t}{4380}\pi\right)\right)\,\mathrm{d}t\\ &=\frac{1}{24}\int_0^{24}400\,\mathrm{d}t + \frac{50}{24}\int_0^{24}\cos\left(\frac{t}{12}\pi\right)\,\mathrm{d}t+ \frac{200}{24}\int_0^{24}\cos\left(\frac{t}{4380}\pi\right)\,\mathrm{d}t \\ \text{Note $t=0$ to $t=24$ is one complete period for the integrand in red, so the red integral will evaluate to zero. However, $t=0$ to $t=24$ is \textit{less} than one cycle for the integrand in blue, so we expect this will contribute some non-zero quantity to the average.} \\ &=400+ 0 + \frac{200}{24}\left[\frac{4380}{\pi}\sin\left(\frac{t}{4380}\pi\right)\right]_0^{24}\\ &=400+\frac{25}{3}\cdot\frac{4380}{\pi}\sin\left(\frac{24}{4380}\pi\right)\\ &=400+\frac{25}{3}\cdot\frac{4380}{\pi}\sin\left(\frac{2}{365}\pi\right)\\ &\approx 400+199.99\\ &=599.99 \quad\text{ ppm} \end{aligned}$$

Remark: $C(0) = 400+ 50+ 200$. The red term comes from the daily variation, and over the first day this will have an average of 0. The blue term comes from the seasonal variation, which changes dramatically over the course of an entire year but won't change very much over the course of a single day. So, it is reasonable that the average concentration over the first day should be close to (but not exactly) $400+200$ ppm.

(c) The average of $N(t)$ over $[0,8760]$ is:

$$\begin{aligned} \frac{1}{8760}\int_0^{8760} \left ( 350+200\cos\left(\frac{t}{4380}\pi\right)\right)\,\mathrm{d}t&=350 + \frac{200}{8760}\bigg[\frac{4380}{\pi}\sin\left(\frac{t}{4380}\pi\right)\bigg]_0^{8760} \\&=350+\frac{200}{8760}\left[\frac{4380}{\pi}\sin\left(\frac{8760}{4380}\pi\right)\right]\\ &=350+\frac{100}{\pi}\sin\left(2\pi\right)\\&=350 \end{aligned}$$

Since the average of $C(t)$ was 400, this gives us an absolute error of $|400-350|=50$ ppm, for a relative error of $$\frac{50}{400} = 0.125,$$ or 12.5%.

That is: sampling at the same time every day, rather than throughout the day, lead to an error of 12.5% in the yearly average concentration of carbon dioxide.

The question tells you $\frac{1}{1-0}\int_0^1 f(x)\,\mathrm{d}x = \frac{f(0)+f(1)}{2}$. Note $f(0)=c$, and $f(1) = a+b+c$.

$$\begin{aligned} \frac{1}{1-0}\int_0^1 f(x)\,\mathrm{d}x &= \frac{f(0)+f(1)}{2} = \frac{c+(a+b+c)}{2}\\ \int_0^1 \big(ax^2+bx+c \big)\,\mathrm{d}x&=\frac{a+b+2c}{2}\\ \left[\frac{a}{3}x^3+\frac{b}{2}x^2+cx \right]_0^1&=\frac{a}{2}+\frac{b}{2}+c \\\frac{a}{3}+\frac{b}{2}+c&=\frac{a}{2}+\frac{b}{2}+c\\ \frac{a}{3}&=a\\ a&=0 \end{aligned}$$

That is, $f(x)$ is linear.

The function $g(x)=f(a+b-x)$, on the interval $[a,b]$, is a mirror of the function $f(x)$, with $g(a)=f(b)$ and $g(b)=f(a)$. So, $\int_a^b f(a+b-x)\,\mathrm{d}x = \int_a^bf(x)\,\mathrm{d}x$, and hence $\frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x = \frac{1}{b-a}\int_a^bf(x)\,\mathrm{d}x$, so the average value of $f(a+b-x)$ on $[a,b]$ is $A$.

Alternately, we can evaluate $\frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x$ directly, using the substitution $u=a+b-x$, $\mathrm{d}x=-\mathrm{d}x$:

$$\begin{aligned} \frac{1}{b-a}\int_a^b f(a+b-x)\,\mathrm{d}x&=\frac{-1}{b-a}\int_{u(a)}^{u(b)} f(u)\,\mathrm{d}u\\ &=\frac{-1}{b-a}\int_{b}^{a} f(u)\,\mathrm{d}u \\&=\frac{1}{b-a}\int_{a}^{b} f(u)\,\mathrm{d}u \\&=A \end{aligned}$$